Electrical Resistance Calculation Help

[drgeoff]

I have run a number of tests and here are some of the results...

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v1		v2		r1	r2	r2(cal)		r1(16.8k fix)	r2(16.8k fix)	vd		v2*r1fix
2.154712	3.264191	1000	2200	2776.807387	943.8202247	2382.940343	1.109479	3080.809483

You don't appear to be fully understanding the discussion.

1. There is no fix required for R1.

2. You show no figures with R2 absent nor any attempt to confirm the actual value of the input resistance of the ADC channels.

[davidcoton]

According to https://www.khanacademy.org/science/ele ... ge-divider I should be using this formula to find r2:
r2 = (-v1*r1)/(v1-v2)

I haven't checked the source (no time, have to go to work) but that equation gives a negative resistance. Unlikely.

To correct for the error, I believe I am going to have to take a constant and add it to one of the equations. Is the constant I want to add K? for instance, 1.59 or 2.4? Basically for calibration?

The errors discussed so far do not require a correction added anywhere -- except that you must get the right values to apply in the corrections. Such an additon is known as a "fiddle factor" and is a bad idea unless you can justify it from basic theory.
The "K" I have use in the contuctivity equation accounts for the geometry of you probes and test cell. It is a multiplier not an addition.

I think you need to take a step back, read @drgeoff's and my replies again, and again until you understand what we are saying and why. It's not all easy or intuitive, you may need to find someone local with a scientific/maths background and sit over coffee to discuss it.

[vvarrior]

You don't appear to be fully understanding the discussion.

1. There is no fix required for R1.

2. You show no figures with R2 absent nor any attempt to confirm the actual value of the input resistance of the ADC channels.

1. I did read that r1 may not need a fix and have done the calculation with and without it. Without the fix resulted in more accurate results.

2. But now I know what I need to do: confirm the results of input resistance of ADC. But according to davidcoton that number is irrelevant anyway. This may be some of my confusion.
If I know the values of v1, v2, r1 and r2, shouldn't I be able to calculate the values of rL in this circuit?

rL.jpg (10.66 KiB) Viewed 1447 times

I would like to verify the resistance of the ADC channel, however, I don't know how to do that. I will contact AB electronics and see what they suggest.

And just to clarify, you say the board has 16.8k resistance in parallel, but because the vout on the board is between 10k and 6.8k, wouldn't the correction number I want to use 10k instead of 16.8k?

I haven't checked the source (no time, have to go to work) but that equation gives a negative resistance. Unlikely.

Here is how I came up with that formula and it seems to work:
https://www.khanacademy.org/science/ele ... ge-divider
v2 = v1*(R2/(R2+R1))
https://www.mathpapa.com/algebra-calculator.html solved for R2 instead of v1 and this is the equation:
r2 = (-v1*r1)/(v1-v2)
Which results in a positive r2.

Thank you both for all of your input and knowledge. You have been an enormous help in understanding this.

[drgeoff]

The easiest way to measure RL is to read the two voltages you get with R1 of say 10k and R2 absent. Then the potential divider is just R1 and RL. The ratio of V1 to V2 will be RL/(R1+RL). Easily solved for RL.

If you look at the schematic you will see that each input channel goes to a 10k resistor. The other end of that goes to the ADC chip and a 6.8k resistor to ground. So if the chip were to take all the current the input resistance of the channel would be 10K. If the chip takes no current the input resistance would be 10k + 6.8k = 16.8k. Clearly it is going to be some value between those two extremes.

Once you know the actual value of R, you can get a much more accurate value for R2 or the water. You measure V1 and V2. You use them and the known value of R1 to calculate the effective resistance of the parallel pair of R2 and RL. Call it RM. I already gave you the formula for resistors in parallel. In this case RM = R2 x RL/(R2 + RL). Solve for the final unknown R2.

[drgeoff]

Here is how I came up with that formula and it seems to work:
https://www.khanacademy.org/science/ele ... ge-divider
v2 = v1*(R2/(R2+R1))
https://www.mathpapa.com/algebra-calculator.html solved for R2 instead of v1 and this is the equation:
r2 = (-v1*r1)/(v1-v2)
Which results in a positive r2.

Don't know why you think that results in a positive r2. v1 is greater than v2. The top is negative, the bottom is positive. Result is negative as davidcoton said.

[drgeoff]

And just to clarify, you say the board has 16.8k resistance in parallel,

Further evidence that you are not paying sufficient attention. What I said was "Each channel presents a load of no more than 16,800 ohms." I then used 16,800 to calculate and show you the minimum error that the loading by the ADC board is introducing. The true figure for the board is less than 16,800 so the error will be even more than I showed.

[vvarrior]

Please excuse my ignorance in this field. I am a paramedic by trade and this is a hobby that I am learning. I don't mean to come off knowing it all and I truly want to understand what's going on underneath. Here are the results of a test run with the setup in the picture. Please note v1 goes to ADC ch 4 (Vout) and v2 goes to ADC ch 5 (Vin). Test results:
v1 = 0.000000
v2 = 3.289171
r1 = 10,000 ohm
r2 = jumper

blank adc with circ - jumper.jpg (114.19 KiB) Viewed 1412 times

This means that the ohms on r2 are 0 correct?

It seems this analog to digital converter board is designed as a single ended input. The proper ADC board to do this is an ADC Differential Pi. AB Electronics informed me that I could remove the resistors on the board and drill out a ground to make it differential. Thoughts?

I apologize for making a simple problem more complex.

[davidcoton]

Sorry I don't have time to write anything in enough detail to be helpful. At present I am not sure where and why your understanding is going awry -- which is why I suggested you need some face to face help.

I don't think a "differetial mode" ADC will help (it would mean you could get the voltage across R1 directly, instead of calculating V1-V2), but that is not the major problem. It is possible that differetial mode would improve the input resistance issue, but also it might not.

Testing with R2=0 (link) will not tell you much. It needs to be infinite (not present) to measure the input resistance of the ADC.

But now I know what I need to do: confirm the results of input resistance of ADC. But according to davidcoton that number is irrelevant anyway. This may be some of my confusion.

Not at all. The input resistance Rk is vital to an accurate calculation of R2, the cell resistance. I said that you do not need to correct the voltages (if you did need to, that calculation would also need Rk)

[vvarrior]

1. The effect of 16,800 ohms is that a 1,000 ohm resistor becomes 943.8 and a 2200 ohm one appears to be 1945.3.

You can see the effect of and measure the ADC input resistance by omitting R2. If the ADC channel was infinite resistance then v1 and v2 would be equal. Use a 10,000 ohm for R1, leave out R2 and I would expect v2 to be about 2 volts.

I did as suggested and reread this entire thread multiple times, following more logic. drgeoff, you are correct, when I remove r2 entirely (no jumper or infinite resistance as you said), Vout (v1) is 2.070157v and Vin (v2) 3.295735 resulting in 26891.27089 ohms. I am assuming this is the 10,000 r1 plus the 16800 resistors on the ADC channel.

Some tests:

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v1		v2		r1		r2 (calc)
3.29504		2.070852	10000		26916.12726
3.295696	2.07008		10000		26890.11893
3.296314	1.740202	15000		31774.5188

The difference of the r2 and r1 is indeed very close to 16800! Which made everything else make sense. Then I used the formulas:
r2m=(v2*r1)/(v1-v2)
rk = 16800
r2 = (r2m*rk)/(r2m-rk)
The results are now highly accurate.

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v1		v2		r1	r2	r2 calc
3.293225	0.377445	15000	2200	2,195
3.293225	0.194591	15000	1000	998
3.295696	1.403528	15000	33000	32,946
3.292608	0.069188	15000	330	328

That was a tough few days. Thank you for being patient with me gentlemen. Success and time for the next part.

[davidcoton]

I did as suggested and reread this entire thread multiple times, following more logic. drgeoff, you are correct, when I remove r2 entirely (no jumper or infinite resistance as you said), Vout (v1) is 2.070157v and Vin (v2) 3.295735 resulting in 26891.27089 ohms. I am assuming this is the 10,000 r1 plus the 16800 resistors on the ADC channel.

As they used to say in school, show your working. It does look as though the result you quote is R1+Rk, but we have no way of knowing. The formulae for R2 @drgeoff and I gave would give Rk directly.

[drgeoff]

There is another way of looking at this which is possibly more direct. Uses both Ohm's Law and Kirchoff's Current Law. The latter states that at any point in a circuit the sum of all the currents arriving equals the sum of all the currents leaving. In this case consider the point where R1, R2 and RL join. Kirchoff says that the current through R1 equals the current through R2 plus the current through RL. Nothing earth-shattering about that.

The current though R1 can be found from Ohm's Law because the resistance of R1 is known and the difference in the two measurements gives the voltage across it. We can also apply Ohm's Law to RL because we now know its resistance and we measure the voltage across it so the current can be calculated. Now that we have two of the three currents, the remaining one through R2 is the difference. Now we have the current through R2 and we measured the voltage across it. Ohm's Law gives is resistance.

[vvarrior]

This missing resistors and points of measure were the missing link for my understanding. Thank you for sharing your knowledge. Now to figure out calculating EC from ohms.